Simplify and expand the following expression: $ \dfrac{3}{p + 3}+ \dfrac{3}{3p - 18}- \dfrac{p}{p^2 - 3p - 18} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{3}{3p - 18} = \dfrac{3}{3(p - 6)}$ We can factor the quadratic in the third term: $ \dfrac{p}{p^2 - 3p - 18} = \dfrac{p}{(p + 3)(p - 6)}$ Now we have: $ \dfrac{3}{p + 3}+ \dfrac{3}{3(p - 6)}- \dfrac{p}{(p + 3)(p - 6)} $ The least common multiple of the denominators is: $ (p + 3)(p - 6)$ In order to get the first term over $(p + 3)(p - 6)$ , multiply by $\dfrac{3(p - 6)}{3(p - 6)}$ $ \dfrac{3}{p + 3} \times \dfrac{3(p - 6)}{3(p - 6)} = \dfrac{9(p - 6)}{(p + 3)(p - 6)} $ In order to get the second term over $(p + 3)(p - 6)$ , multiply by $\dfrac{p + 3}{p + 3}$ $ \dfrac{3}{3(p - 6)} \times \dfrac{p + 3}{p + 3} = \dfrac{3(p + 3)}{(p + 3)(p - 6)} $ In order to get the third term over $(p + 3)(p - 6)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{p}{(p + 3)(p - 6)} \times \dfrac{3}{3} = \dfrac{3p}{(p + 3)(p - 6)} $ Now we have: $ \dfrac{9(p - 6)}{(p + 3)(p - 6)} + \dfrac{3(p + 3)}{(p + 3)(p - 6)} - \dfrac{3p}{(p + 3)(p - 6)} $ $ = \dfrac{ 9(p - 6) + 3(p + 3) - 3p} {(p + 3)(p - 6)} $ Expand: $ = \dfrac{9p - 54 + 3p + 9 - 3p}{3p^2 - 9p - 54} $ $ = \dfrac{9p - 45}{3p^2 - 9p - 54}$ Simplify: $ = \dfrac{3p - 15}{p^2 - 3p - 18}$